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A matrix $A$ is called idempotent if $A^2 = A$. I am just wondering if such matrix can be complex. Anyone can help give an example or proof that it has to be real? Thanks!

Tuyet
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    There is nothing the relation $A^2=A$ that suggests entries have to be real (for me they could be polynomials, quaternions or whatever ring you like) so maybe you could share your thoughts on why you thought this implies "real". Maybe you were thinking of eigenvalues (which can only be $0$ or $1$, both of which are indeed real)? – Marc van Leeuwen Mar 30 '17 at 12:23
  • Such $A$ is similar to a real diagonal matrix. – Jonas Meyer Mar 30 '17 at 18:18

3 Answers3

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A assume that by "can $A$ be complex", you mean "can $A$ have any non-real entries". Well, it can! For instance, take $$ A = \pmatrix{1&i\\0&0} $$ In general: for any complex column-vector $x$, $A = \frac{xx^*}{x^*x}$ (where $*$ denotes the conjugate-transpose) is such a matrix.

Joonas Ilmavirta
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Ben Grossmann
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A projection to a subspace is idempotent. Therefore $A$ has no reason to be real. For example, take a subspace $S$ of $\mathbb{C}^2$ and $A$ be the matrix of the projection on to $S$ with respect to the standard basis.

Quang Hoang
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    I can see why this question was asked though. Complex numbers are usually associated with rotation, and you'd naturally think a rotation can't be idempotent. Maybe the question meant to ask about complex eigenvalues or something? – user541686 Mar 30 '17 at 04:03
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Any matrix $A = \pmatrix{a&b\\c&1-a}$ will be idempotent provided that $a^2+bc=a$

Tyberius
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